Reputation: 559
I am doing the next query:
SELECT id, name, keyt
FROM table
WHERE id = (SELECT t2.id FROM table t2 WHERE t2.keyt=21 ORDER BY RAND() LIMIT 1)
Supposing table is like this:
| id | name | keyt |
+ ------------------------- +
| 1 | Hello | 21 |
| 3 | Katzet | 1 |
| 1 | Welcome | 1 |
| 2 | Two | 21 |
| 2 | Other | 1 |
It should return one of this pairs:
So, the idea is:
Get one id, which has the keyt
value set to 21
Then, get all the rows with this selected id (independently of all the other keyt values)
If I do as you suggested... I would get mixed id values, and all result rows must have the same id.
Upvotes: 1
Views: 558
Reputation: 915
Your table has two 21 in the keyt column so your subquery in the where clause returns 2 values if id that is 1 and 2.So what you need to do is instead of using an equal to operator "=" use IN operator in the where clause.
SELECT id, name, keyt FROM table WHERE id IN (SELECT t2.id FROM table t2 WHERE t2.keyt=21 ORDER BY RAND())
Upvotes: 0
Reputation: 33945
SELECT x.*
FROM my_table x
JOIN
( SELECT id
FROM my_table
WHERE keyt = 21
ORDER
BY RAND() LIMIT 1
) y
ON y.id = x.id;
Upvotes: 1
Reputation: 3889
The subquery in this query
SELECT id, name, keyt
FROM table
WHERE id = (SELECT t2.id FROM table t2 WHERE t2.keyt=21 ORDER BY RAND() LIMIT 1)
would return only one record as it has LIMIT 1
added at the end.
Also, in your question, the table contains only 1 record for which value of
keyt
= 21, due to which you're getting only one record.
If you want more records, you should remove the LIMIT
. In that case you may rephrase your query as:
SELECT id, name, keyt
FROM table
WHERE id IN (SELECT t2.id FROM table t2 WHERE t2.keyt=21 ORDER BY RAND())
Hope this is what you expected. As your actual goal is not very clear from the question.
Upvotes: 1